Proving a series is convergent - $\sum _{n=1} ^\infty \frac{(-1)^n}{n}$ without using alternating series test

Question

$$\sum _{k=1} ^\infty \frac{(-1)^k}{k}$$

I know this question has been answered a few times but my professor has not taught alternating series test yet or anything other than ratio test, root test and comparison test where $a_i \geq 0$ for every $i \in \mathbb N$ and $\sum_{i=1} ^\infty a_i$ converges and if $|b_i| \leq a_i$ for every i then $\sum_{i=1} ^\infty b_i$ converges absolutely.

So here's my attempt using Cauchy criterion.

What we know: We say that the series $\sum _{i=1}^\infty a_i$ converges if the sequence of partial sums $(S_i)_i$$_\in$$_\mathbb N$ converges.

From Cauchy criterion, $(S_i)_i$$_\in$$_\mathbb N$ converges if and only if it is a Cauchy sequence.

It is quite obvious that $$\lim_{k\to\infty} S_k = \sum _{k=1}^\infty \frac{(-1)^k}{k}$$.

I denote $\lim_{k\to\infty} S_k = S$

Suppose ($S_k$) is convergent. Then $\forall \epsilon \gt 0, \exists N \in \mathbb N$ such that $\forall n \geq N,$

$|S_n - S|$ = $\vert \sum_{k=n+1} ^\infty \vert$ $\lt \epsilon$

I am stuck here. Is it possible to find such N for all $\epsilon \gt 0$ to hold

$\vert \sum_{k=n+1} ^\infty \vert$ $\lt \epsilon$

to be true?

(If yes, then the sequence ($S_k$) is convergent so $\sum_{k=1} ^\infty \frac{(-1)^k}{k}$ is convergent by the definition but I don't quite understand if we can always find such N)

edit: I don't think ratio test or root test are applicable to solve this and is the alternating series test the only way to solve this problem?

Answer 1

Hint. Let $$ s_n=\sum_{k=1}^n\frac{(-1)^k}{k} $$ Then observe that $$ s_1<s_3<\cdots<s_{2n-1}<s_{2n+1}<s_{2n+2}<s_{2n}<s_{2n-2}<\cdots<s_4<s_2 $$ Hence $a_n=s_{2n-1}$ is increasing and upper bounded, by $s_2$, while $b_n=s_{2n}$ is decreasing and lower bounded by $s_1$. Hence both converge, and since $a_n<b_n$, then $$ \lim a_n\le \lim b_n $$ But $b_n-a_n=\frac{1}{2n}\to 0$, and hence $$ \lim a_n= \lim b_n $$

Note. Inevitably, the idea of the proof of Alternating series test is used in the above proof.

Answer 2

Not only can we show that the series of interest converges, we can also evaluate it using elementary tools only. We now present two approaches.

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METHODOLOGY $1$:

Noting that $\int_0^1 x^{n-1}\,dx=\frac1n$, we can write

$$\begin{align} \sum_{n=1}^N \frac{(-1)^n}{n}&=\sum_{n=1}^N (-1)^n\int_0^1 x^{n-1}\,dx\\\\ &=-\int_0^1 \sum_{n=1}^N (-x)^{n-1}\,dx\\\\ &=-\int_0^1 \frac{1-(-x)^N}{1+x}\,dx\\\\ &=-\log(2)+(-1)^N\int_0^1 \frac{x^N}{1+x}\,dx \end{align}$$

Since $\left|\frac{x^N}{1+x}\right|\le x^N$, we have the estimate

$$\left|\int_0^1 \frac{x^N}{1+x}\,dx\right|\le \frac{1}{N+1}$$

Therefore,

$$\lim_{N\to\infty }\sum_{n=1}^N \frac{(-1)^n}{n}=-\log(2)$$

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METHODOLOGY $2$:

$$\begin{align} \sum_{n=1}^{2N}\frac{(-1)^n}{n}&=\sum_{n=1}^N \frac1{2n}-\sum_{n=1}^N \frac1{2n-1}\\\\ &=\sum_{n=1}^N \frac1{2n}-\left(\sum_{n=1}^{2N}\frac1n-\sum_{n=1}^N\frac1{2n}\right)\\\\ &=-\sum_{n=N+1}^{2N}\frac1n\\\\ &=-\sum_{n=1}^N \frac{1}{n+N}\\\\ &=-\frac1N \sum_{n=1}^N\frac1{1+(n/N)} \end{align}$$

The last expression is the Riemann Sum for $-\int_0^1 \frac1{1+x}\,dx=-\log(2)$ as expected.

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If these methodologies are not quite the way forward that the OP is seeking, then we simply note

$$\begin{align} \left|\sum_{n=1}^{2N}\frac{(-1)^n}{n}\right|&=\left|\sum_{n=1}^N \frac1{2n}-\sum_{n=1}^N \frac1{2n-1}\right|\\\\ &=\sum_{n=1}^N \frac{1}{2n(2n-1)}\\\\ &\le \frac12 \sum_{n=1}^N \frac1{n^2} \end{align}$$

and the series of interest converges by comparison with the series $\sum_{n=1}^\infty \frac{1}{n^2}$ (Note that $\sum_{n=2}^\infty \frac{1}{n^2}\le \sum_{n=2}^\infty \frac{1}{n(n-1)}=1$).

Answer 3

We can process the terms in pairs because two successive partial sums differ in $\pm\dfrac1n$, which tends to zero.

Now let us study the series with general term

$$\frac1{2n}-\frac1{2n+1}=\frac1{(2n+1)2n}\le\frac1{4n^2}.$$

For this upper bound, we will group the second and third terms, then the fourth to seventh, and so on, each time doubling the length.

$$4S=1+\left(\frac1{2^2}+\frac1{3^2}\right)+\left(\frac1{4^2}+\frac1{5^2}+\frac1{6^2}+\frac1{7^2}\right)+\cdots \\\le1+\left(\frac1{2^2}+\frac1{2^2}\right)+\left(\frac1{4^2}+\frac1{4^2}+\frac1{4^2}+\frac1{4^2}\right)+\cdots \\=1+\frac12+\frac14+\cdots,$$ a well known convergent series.